%I #13 Aug 23 2015 04:46:02
%S 1,1,10,2896,3941598,15277017432,135277939358160,2374127830286012160,
%T 74701932179186551241520,3911393168902074440088524160,
%U 321715999535364496261149134365440,39702971502659332476270701578180454400,7081620512071831837127802029303335215878400
%N Number of distinct n-colored necklaces with 4 beads per color.
%H Alois P. Heinz, <a href="/A208185/b208185.txt">Table of n, a(n) for n = 0..50</a>
%F a(n) = Sum_{d|4} phi(4/d)*(n*d)!/(d!^n*n*4) if n>0 and a(0) = 1.
%F For n > 0, a(n) = (4*n)!/(4*n*24^n) + (n-1)!/2 + (2*n-1)!/2^(n+1). - _Vaclav Kotesovec_, Aug 23 2015
%e a(0) = 1: the empty necklace.
%e a(1) = 1: {0000}.
%e a(2) = 10: {00001111, 00010111, 00100111, 01000111, 00011011, 00110011, 00101011, 01010011, 01001011, 01010101}.
%p with(numtheory);
%p a:= n-> `if`(n=0, 1, add(phi(4/d) *(n*d)!/(d!^n *4*n), d={1,2,4})):
%p seq(a(n), n=0..15);
%t Flatten[{1, Table[(4*n)!/(4*n*24^n) + (n-1)!/2 + (2*n-1)!/2^(n+1), {n, 1, 15}]}] (* _Vaclav Kotesovec_, Aug 23 2015 *)
%Y Row n=4 of A208183.
%Y Cf. A000010, A000142.
%K nonn
%O 0,3
%A _Alois P. Heinz_, Feb 24 2012